When using shell elements in finite element simulations, it is crucial to extract the section internal force of the shell element. This article explains how to extract the internal force in ANSYS APDL.
First, let’s talk about the output of stress and internal force of shell elements.
The output items of stress and internal force for thin shell elements and moderately thick shell elements are different. For thin shell elements such as SHELL63, minor stresses (τxz, τyz) and internal forces (Nx, Ny) are not output, whereas moderately thick plate shell elements output these stresses and internal forces.
Note that the internal force output of shell elements is relative to the element coordinate system, and the internal forces on each side of the element are the same, which are the internal forces per unit length of the element. For example, the unit of Mx is “force × length/length”, and the total bending moment of the element can be obtained by multiplying the length of the element. The internal force of the element can be output through the element table. For example, the result output of shell181 is shown in the figure below, and the element table options are as follows:
The above method is for a single element. However, in actual calculation processes, we often need to obtain the total internal force of a certain section, which can be obtained by calculation. Generally, there are two ways: the path integration method and the summation of element node forces method. I personally recommend using the summation of element node forces method, which is simple and fast.
The summation of element node forces method requires mastering two commands: Spoint and Fsum.
Spoint, node, x, y, z
This command defines the position point for force summation. If the summation is not located under the overall Cartesian coordinate system, node definition or the Rsys command can be used.
Fsum, lab, Item
This command calculates the resultant force and moment of all the nodal forces of the selected element set and node set. Therefore, when calculating the internal force of a specific section, the elements and nodes adjacent to the section should be selected.
Below is an example of a cantilever plate to illustrate the basic idea.
A certain concrete cantilever plate with a thickness of 100mm and dimensions of 900mm X 2000mm, with a concrete grade of C30, is subjected to a uniform load of 0.5KN/m^2 within 100mm range from the end of the plate. The shear force and bending moment of the mid-span section of the plate are required.
Theoretical solution: Fz=0.5*0.1*2=0.1KN
The APDL code:
finish /clear /prep7 et,1,shell181 r,1,100 mp,dens,1,2600e-12 mp,prxy,1,0.3 mp,ex,1,3.0e4 blc4,,,900,2000 wpoffs,450 wprota,,,90 asbw,all wpoffs,,,350 asbw,all wpcsys,-1 asel,s,loc,x,800,900 sfa,all,1,pres,-0.5e-3 allsel,all esize,20 amesh,all /solu lsel,s,loc,x,0 dl,all,,all,0 outres,all,all allsel,all solve /post1 nsel,s,loc,x,450 esln,s,0 esel,u,cent,x,0,450 nsel,s,loc,x,450 spoint,,450,1000 fsum
The solution results are consistent with the theoretical results!